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\title{\heiti\zihao{2} 习题5.1}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{填空:}
\subsection{$\mathrm{d}(\ln x )=\frac{\mathrm{~d} x}{x}$}


\subsection{$\mathrm{d}(\frac{1}{2}\ln(1+2x))=\frac{\mathrm{~d} x}{1+2 x}$}


\subsection{$\mathrm{d}(2\sqrt x)=\frac{\mathrm{~d} x}{\sqrt{x}}$}


\subsection{$\mathrm{d}(x^2+x)=(2 x+1) \mathrm{~d} x$}


\subsection{$\mathrm{d}(\sin x - \cos x)=(\cos x+\sin x) \mathrm{~d} x$}


\subsection{$\mathrm{d}(\ln(x+\sqrt{1+x^2}+C))=\frac{\mathrm{~d} x}{\sqrt{1+x^{2}}}$}


\subsection{$\mathrm{d}(-\frac{1}{a}e^{-ax})=\mathrm{e}^{-a x} \mathrm{~d} x$}


\subsection{$\mathrm{d}(-\frac{1}{4}\cos 2x)=\cos x \sin x \mathrm{~d} x$}


\subsection{$\mathrm{d}(\ln\ln x)=\frac{\mathrm{~d} x}{x \ln x}$}


\subsection{$\mathrm{d}(\sqrt{x^2+a^2})=\frac{x \mathrm{~d} x}{\sqrt{x^{2}+a^{2}}}$}


\subsection{$\mathrm{d}(-\frac{1}{3}\cos^3x)=\cos ^{2} x \sin x \mathrm{~d} x$}


\subsection{$\mathrm{d}(\frac{x}{2}-\frac{\sin 2x}{4})=\sin ^{2} x \mathrm{~d} x$}

\section{求下列函数的一阶和二阶微分:}
\subsection{$y=x^{4}+5 x$}
\textbf{解}\quad
$$\mathrm{~d} y=\left(4 x^{3}+5\right) \mathrm{~d} x 
,d^{2} y=\left(12 x^{2}\right) \mathrm{~d} x^{2}$$

\subsection{$y=\cos n x$}
\textbf{解}\quad
$$\mathrm{~d} y=(-n \sin n x) \mathrm{~d} x, d^{2} y=\left(-n^{2} \cos n x\right) \mathrm{~d} x^{2}$$

\subsection{$y=x \ln x$}
\textbf{解}\quad
$$\mathrm{~d} y=(\ln x+1) \mathrm{~d} x,d^{2} y=\left(\frac{1}{x}\right) \mathrm{~d} x^{2}$$

\subsection{$y=\sqrt{1+t^{2}}+1$}
\textbf{解}\quad
$$\mathrm{~d} y=\frac{t}{\sqrt{1+t^{2}}} d t, d^{2} y=\frac{1}{\left(1+t^{2}\right)^{\frac{3}{2}}}dt^2$$

\subsection{$y=\frac{u+1}{u-1}$}
\textbf{解}\quad
$$\mathrm{~d} y=\frac{-2}{(n-1)^{2}} d u, u^{2} y=4(u-1)^{-3} d u^2$$

\subsection{$y=(1+2 r)^{-5}$}
\textbf{解}\quad
$$\mathrm{~d} y=-10(1+2 r)^{-6} d r, d^{2} y=120(1+2 r)^{-7} d r^{2}$$

\subsection{$y=\sqrt{x+\ln x}$}
\textbf{解}\quad
$$\mathrm{~d} y=\frac{1+x}{2 x \sqrt{x+\ln x}} \mathrm{~d} x, d^{2} y=\frac{-x^{2}-4 x-2 \ln x-1}{4 x^{2}(x+\ln x) \sqrt{x+\ln x}} \mathrm{~d} x^{2}$$

\subsection{$y=e^{-\frac{x^{2}}{4}}$}
\textbf{解}\quad
$$\mathrm{~d} y=-\frac{x}{2} e^{-\frac{x^{2}}{4}} \mathrm{~d} x, d^{2} y=-\frac{1}{2} e^{-\frac{x^{2}}{4}}\left(1-\frac{x^{2}}{2}\right) \mathrm{~d} x^{2}$$

\subsection{$y=x^{2} \mathrm{e}^{3 x}$}
\textbf{解}\quad
$$\mathrm{~d} y=\left(2 x+3 x^{2}\right) e^{3 x} \mathrm{~d} x, d^{2} y=e^{3 x}\left(2+12 x+9 x^{2}\right) \mathrm{~d} x^{2}$$

\subsection{$y=(3 x-2) \sin 2 x$}
\textbf{解}\quad
$$\mathrm{~d} y=(3 \sin 2 x+2(3 x-2) \cos 2 x) \mathrm{~d} x,d^{2} y=(-12 x \sin 2 x+12 \cos 2 x+8 \sin 2 x) \mathrm{~d} x^{2}$$

\subsection{$y=\sin x^{2}$.}
\textbf{解}\quad
$$\mathrm{~d} y=2 x \cos x^{2} \mathrm{~d} x, d^{2} y=\left(2 \cos x^{2}-4 x^{2} \sin x^{2}\right) \mathrm{~d} x^{2}$$

\section{根据下面给出的 $x$ 和 $\mathrm{d} x$ 的值,求微分 $\mathrm{~d} y$ :}
\subsection{$y=x^{2}+5 x, x=3, \mathrm{~d} x=\frac{1}{2}$}
\textbf{解}\quad
$$\mathrm{~d} y=(2 x+5) \mathrm{~d} x=\frac{11}{2}$$

\subsection{$y=\sqrt{4+5 x}, x=0,$ d $x=0.004$}
\textbf{解}\quad
$$\mathrm{~d} y=\frac{5}{2 \sqrt{4+5 x}} \mathrm{~d} x=0.005$$

\subsection{$y=\frac{1}{x+1}, x=1, \mathrm{~d} x=-0.001$}
\textbf{解}\quad
$$\mathrm{~d} y=-\frac{1}{(x+1)^{2}} \mathrm{~d} x=0.00025$$

\subsection{$y=\tan x, x=\frac{\pi}{4}, \mathrm{~d} x=-0.1$}
\textbf{解}\quad
$$\mathrm{~d} y=\frac{1}{\cos ^{2} x} \mathrm{~d} x=-0.2$$

\section{假设我们不知道 $g(x)$ 的表达式,但知道 $g(2)=-4$ 和 $\mathrm{g}^{\prime}(x)=\sqrt{x^{2}+5}$.}
\subsection{利用线性近似估计 $g ( 1.96 )$ 和 $g(2.05)$}
\textbf{解}\quad
$\mathrm{~d} y=\sqrt{x^{2}+5} \mathrm{~d} x$,所以$g(1.96)=-4.12,g(2.05)=-3.85$

\subsection{在4.1中.你给出的估计是偏大还是偏小? 请解释原因.}
\textbf{解}\quad
$g^{\prime\prime}(x)$在$x>0$时大于$0$.其为凸函数.所以估计偏小.



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\section{}
\textbf{解}\quad

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\textbf{证}\quad

\textbf{\textcolor{red}{注}}\quad